Advertisements
Advertisements
प्रश्न
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
Advertisements
उत्तर
`lim_(x -> 0) (x(6^x - 3^x))/(cos (6x) - cos (4x))`
= `lim_(x -> 0) (x*3^x (2^x - 1))/(-2sin ((6x + 4x)/2) sin((6x - 4x)/2))`
= `1/(-2) lim_(x -> 0) (x*3^x (2^x - 1))/(sin 5x*sinx)`
= `1/(-2) lim_(x -> 0) ((x*3^x (2^x - 1))/x^2)/((sin 5x* sinx)/x^2) ...[("Divide Numerator and"),("Denominator by" x^2),(∵ x -> 0"," ∴ x ≠ 0 ∴ x^2 ≠ 0)]`
= `1/(-2) (lim_(x -> 0) [3^x ((2^x - 1))/x])/(lim_(x -> 0) ((sin5x)/x * sinx/x))`
= `1/(-2) (lim_(x -> 0)(3x) * lim_(x -> 0) ((2^x - 1)/x))/(lim_(x -> 0) (sinx/(5x) xx 5) * lim_(x -> 0) (sinx/x))`
= `1/(-2) xx (3^circ xx log 2)/(1 xx 5 xx 1) ...[(because x -> 0"," 5x -> 0),(lim_(x -> 0) ("a"^x - 1)/x = log "a"),(lim_(x -> 0) sinx/x = 1)]`
= `(-1)/10 log 2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x ->0) cos x/(pi - x)`
Evaluate the following limit.
`lim_(x -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
Evaluate the following :
`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
`lim_(x -> 0) |x|/x` is equal to ______.
If f(x) = x sinx, then f" `pi/2` is equal to ______.
Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`
Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`x^(2/3)`
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
`lim_(x -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
`lim_(x -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.
Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
